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Head loss in a pipe—using the Darcy-Weisbach Equation

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## Head loss in a pipe—using the Darcy-Weisbach Equation

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**On noting that**We can write In this case we can calculate the resistance r for a given pipe and use the form to analyses flow in pipe networks Head loss in a pipe—using the Darcy-Weisbach Equation If flow is fully turbulent f will not depend on Reynolds number and r can be considered a constant—i.e. NOT a function of velocity V**Flow in a Parallel Pipe System**So for loop 1 we have For loop 2 Consider flow from A to B through the three pipes in the directions shown. If the flow upstream of A is Q m3/s how is it split between pipes 1, 2 and 3. • The diagram has Two NODES A and B • (points where pipe join) • And Two LOOPS • From A along pipe 1 to B • and back along pipe 2 to A • From A along pipe 2 to B • and back along pipe 3 to A 1 A B 2 3 Head loss will increase as we move in direction of flow and decrease as we move against flow There can be NO net change in head around a closed loop. (1) (2) (3) Continuity IF Q is known we can solve the three equations (Using SOLVER in EXCELL) to obtain Values for the Qi Download excel file “parbal.xls” in notes section**We can extend the ideas to general Pipe Networks exercise**1 in the pipe network lab 4 NODES A, B, C, D 5 Pipes 2 Loops For given Q inputs at A, B, C and D and r’s determine Pipe discharges Loop 1:R1*QQ1*ABS (QQ1) +RR4*QQ4*ABS (QQ4) + RR5*QQ5*ABS (QQ5) = 0 Loop 2: -RR2*QQ2*ABS(QQ2) - RR3*QQ3*ABS(QQ3) + RR4*QQ4*ABS(QQ4) = 0 Balance Node A: QQA + QQ5 – QQ1 = 0 Balance Node B: QQ1 – QQ2 – QQB – QQ4 = 0 Balance Node C: QQ2 – QQ3 – QQC = 0 Balance Node D: QQ3 + QQ4 + QQ5 – QQD = 0 Overall Balance: QQA – QQB – QQC – QQD = 0 More equations than we need—But SOLVER can handle them Note if you “Guess” a wrong direction fro flow the discharge value will be negative**El. 25 m**El. 25 m El. 20 m El. 20 m L L 3 3 L L El. 0 El. 0 2 2 L L P P 1 1 Pipe systems with reservoirs and pumps One Node And ??? Loops Create “pseudo loops” with zero flows (Q = 0) between reservoirs surfaces Then head-loss loop equations are Note: directions around loops There is a positive head loss when we move through a pump opposite to flow direction Continuity**El. 25 m**El. 20 m L 3 L El. 0 2 L P 1 hp h Q More on the pump The efficiency of the pump is also a function of Q It will be zero at “shutoff” and free-delivery and attaint a maximum < 100% for It is important to choose a pump That is efficient for required Q The head provided by the pump is a function of the discharge Q through it The shutoff head the maximum head that can be provided—the pump can lift water to this height BUT water can not flow (Q = 0) The free-delivery This is the maximum flow through the pump. It can only be achieved if no pipe is attached to the pump ( hP = 0).